the position of a body moving along x-axis at time t is given by **x = (t ^{2}-4t+6)** m . The distance travelled by body in time interval t = 0 to t = 3 s is -

(1) 5 m

(2) 7m

(3) 4 m

(4) 3 m

Here x=t^2-4t+6 equation represent the position of that particle along x axis at given time t. Now at t=0 , particle present at x=6 point means x(6,0) then t=1 ,x(3,1); then t=2,x(2,2); and last t=3,x(3,3). Now distance is sclar quantity so along x axis particle move the distance x(6,0) to x(3,1) = 3m, x(3,1) to x(2,2) =1m and last the particle move x(2,2) to x(3,1)=1m . So the distance travelled by body in time interval t=0 to t=3s is (3+1+1)=5m

- 74

As already told by Abu syed.
x at 0 sec = 0^2-4*0+6=6m
x at 1 sec = 3m
x at 2 sec = 2m
x at 3 sec = 3m
Now let's take out distance at particular second.
x at 1st sec = 3m
x at 2nd sec (x at 2 sec - x at 1sec) = 2-3 = 1m (as distance is scalar quantity and hence has no direction therefore no-ve sign)
x at 3rd sec (x at 3 sec - x at 2sec) = 3-2 = 1m.
Now add all the individual distances I.e. 3+1+1 = 5m.

- 110

Abu syed. x at 0 sec = 0^2-4*0+6=6m x at 1 sec = 3m x at 2 sec = 2m x at 3 sec = 3m Now let's take out distance at particular second. x at 1st sec = 3m x at 2nd sec (x at 2 sec - x at 1sec) = 2-3 = 1m (as distance is scalar quantity and hence has no direction therefore no-ve sign) x at 3rd sec (x at 3 sec - x at 2sec) = 3-2 = 1m. Now add all the individual distances I.e. 3+1+1 = 5m.

- 0

X =t - 4 + t b First we have to check where velocity of particle is change v= DX DT = 2+-4=2 (t-2) velocity is changing at t 2 sec motion of particle is like at t=0= X= 6so , magnitude of distance travelled (6-2)=4m at +=3=X=3 so distance travelled during t = 2sec of =X3 -x 2=3-2 =1m so Total distance travelled =(4+1)m = 5m

- 1

The correct option is A.

Given,

x=t2?4t+6

So, the change in velocity is:

dtdx?=2t?4

Since velocity is changing,

At?t=0,x1?=6

At?t=2,x2?=2

So the magnitude of the distance traveled is:

6?2=4m

At?t=3,x3?=3

So distance traveled from?t=2s?to?t=3s

x3??x2?=3?2=1m

Thus the total distance is:

4+1=5m

Given,

x=t2?4t+6

So, the change in velocity is:

dtdx?=2t?4

Since velocity is changing,

At?t=0,x1?=6

At?t=2,x2?=2

So the magnitude of the distance traveled is:

6?2=4m

At?t=3,x3?=3

So distance traveled from?t=2s?to?t=3s

x3??x2?=3?2=1m

Thus the total distance is:

4+1=5m

- 0